An argument for the Rank-Plus-Nullity Theorem

Question

Could you please comment on the legibility of my proof of the Rank-Plus-Nullity Theorem?

Theorem: $A$ is an $(m \times n)$-matrix with $k$ pivot columns. The dimension of the null space of $A$ is $n - k$.

The linear algebra books to which I referred only stated it in the context of linear transformations. So, the verifications of it relied on various propositions about arbitrary vector spaces, but I wanted a demonstration for it in the context of matrices. Here is my demonstration:

$T$ denotes the set of every integer $1 \leq j \leq n$ such that the $j^{\mathrm{th}}$ column vector in $A$ is not a pivot column. $\vert T \vert = n - k$. For each integer $j \in T$, $\boldsymbol{x}_{j} = (x_{i})_{\scriptscriptstyle{i=1}}^{\scriptscriptstyle{n}}$ is the vector in the null space of $A$ defined so that $x_{j} = 1$ and $x_{i} = 0$ for every integer $i \in T\setminus\{j\}$. $X = \{\boldsymbol{x}_{j} \mid j \in T\}$ is a linearly independent set in the null space of $A$. If $X$ spans the null space of $A$, $X$ is a basis for the null space of $A$.

$R$ is the reduced-row echelon matrix that is row equivalent to $A$. The null space of $R$ is the same as the null space as $A$. For any vector $\boldsymbol{y} = (y_{i})_{\scriptscriptstyle{i=1}}^{\scriptscriptstyle{n}}$ in the null space of $A$, \begin{equation*} \boldsymbol{y}^{\prime} = y_{t_{1}}\boldsymbol{x}_{t_{1}} + \ldots + y_{t_{n-k}}\boldsymbol{x}_{t_{n-k}} . \end{equation*} Being a linear combination of vectors in the null space of $A$, $\boldsymbol{y}^{\prime}$ is a vector in the null space of $A$. \begin{equation*} \boldsymbol{y} - \boldsymbol{y}^{\prime} = (z_{i})_{\scriptscriptstyle{i=1}}^{\scriptscriptstyle{n}} \end{equation*} is another vector in the null space of $A$. Moreover, $z_{j} = 0$ for each integer $j \in T$. Since for every integer $1 \leq i \leq n$ that is not in $T$, $z_{i}$ is equal to a linear combination from $\{z_{j} \mid j \in T\}$, $z_{i} = 0$, too. Consequently, $\boldsymbol{y} - \boldsymbol{y}^{\prime} = \boldsymbol{0}_{\scriptscriptstyle{{\mathbb{R}}^{m}}}$, or equivalently, $\boldsymbol{y} = \boldsymbol{y}^{\prime}$. $X$ spans the null space of $A$.

Answer 1

Partial answer$-$I haven't verified your whole argument yet, but here are some comments on its legibility.

It would be more readable if you clarify the kind of statements you are making:

First, if you're assuming something, say "Assume $\ldots$ is $\ldots$" or "Let $\ldots$ be $\ldots$". For example, the statement of your theorem begins with '$A$ is an $\ldots$' When I first read this I thought, "What? Really? Where did $A$ come from?' Similarly for your paragraphs that begin with '$T$ denotes $\ldots$' and '$R$ is the $\ldots$'.

Also, if you're deducing that something is true (especially if it's from the statement immediately before it), say "Therefore..." or "Thus..." etc. Also, what theorems are you applying when? (You don't need to reference every theorem though.)